That the Hankel functions form a basis of solution of Bessel’s equation for any v. The Hankel functions form a basis of solution of Bessel’s equation is J v ( x ) = 1 2 [ H v ( 1 ) + H v ( 2 ) ] _ . Formula used: The Hankel function of order v is H v ( 1 ) = J v ( x ) + i Y v ( x ) , H v ( 2 ) = J v ( x ) − i Y v ( x ) . Calculation: The Hankel function of order v is given as H v ( 1 ) = J v ( x ) + i Y v ( x ) , H v ( 2 ) = J v ( x ) − i Y v ( x ) where H v ( 1 ) and H v ( 2 ) are linearly independent. In order to show that the Hankel function of order v, Therefore, the general solution can be written as y = C 1 H v ( 1 ) + C 2 H v ( 2 ) Let C 1 H v ( 1 ) + C 2 H v ( 2 ) = 0 . Substitute the value of H v ( 1 ) = J v ( x ) + i Y v ( x ) , H v ( 2 ) = J v ( x ) − i Y v ( x ) in C 1 H v ( 1 ) + C 2 H v ( 2 ) = 0 . C 1 ( J v + i Y v ) + C 2 ( J v − i Y v ) = 0 C 1 J v + i C 1 Y v + C 2 J v − C 2 i Y v = 0 Separating the variables in the above equation. C 1 J v + C 2 J v + i C 1 Y v − C 2 i Y v = 0 J v ( C 1 + C 2 ) + ( i C 1 − i C 2 ) Y v = 0 Where H v ( 1 ) and H v ( 2 ) are linearly independent. Therefore, the equation ( C 1 + C 2 ) = 0 and ( C 1 − C 2 ) = 0 . Add the equation ( C 1 + C 2 ) = 0 and ( C 1 − C 2 ) = 0 . ( C 1 + C 2 ) + ( C 1 − C 2 ) = 0 2 C 1 = 0 C 1 = 0 Substitute the value of C 1 = 0 in ( C 1 + C 2 ) = 0 . ( 0 + C 2 ) = 0 C 2 = 0 Hence the solution of the equation is C 1 = 0 , C 2 = 0 . Add the equation H v ( 1 ) = J v ( x ) + i Y v ( x ) and H v ( 2 ) = J v ( x ) − i Y v ( x ) . H v ( 1 ) + H v ( 2 ) = J v ( x ) − i Y v ( x ) + J v ( x ) + i Y v ( x ) 2 J v ( x ) = H v ( 1 ) + H v ( 2 ) J v ( x ) = 1 2 [ H v ( 1 ) + H v ( 2 ) ] Hence, the Hankel functions form a basis of solution of Bessel’s equation is J v ( x ) = 1 2 [ H v ( 1 ) + H v ( 2 ) ] _ .

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